Say we have an equation
$\lambda_{\epsilon}(s)=-\frac{1}{\pi s^2}\ln(1-\epsilon)$
$\forall s\in (0,(M \mathcal{k})^{-\frac{1}{\alpha}})$
where $s$ can be obtained by $s=(M \mathcal{k})^{-\frac{1}{\alpha}}$
How can we simplify the equation
$\lambda_{\epsilon}(s)=-\frac{1}{\pi}(M \mathcal{k})^{\frac{2}{\alpha}}\ln(1-\epsilon)$
into
$\lambda_{\epsilon}(s)=\frac{1}{\pi}(M \mathcal{k})^{\frac{2}{\alpha}}\epsilon + \Theta(\epsilon^2)$
Is it using Sterling approximation to get the asymptotic order expression?
Can someone teach me or give some hints regarding how we can simplify that equation
It's just using a Taylor series expansion of $\ln(1-\epsilon)$.
The Taylor series expansion is
$$\ln(1-x)=-\sum\limits_{n=1}^\infty \frac{x^n}{n}$$
Hence we have
$$\begin{align} \lambda_{\epsilon}(s)&=-\frac{1}{\pi}(M \mathcal{k})^{\frac{2}{\alpha}}\ln(1-\epsilon) \\&=\frac{1}{\pi}(M \mathcal{k})^{\frac{2}{\alpha}}\sum\limits_{n=1}^\infty \frac{\epsilon^n}{n} \\&=\frac{1}{\pi}(M \mathcal{k})^{\frac{2}{\alpha}}\epsilon+\frac{1}{\pi}(M \mathcal{k})^{\frac{2}{\alpha}}\sum\limits_{n=2}^\infty \frac{\epsilon^n}{n} \\&=\frac{1}{\pi}(M \mathcal{k})^{\frac{2}{\alpha}}\epsilon+\epsilon^2\frac{1}{\pi}(M \mathcal{k})^{\frac{2}{\alpha}}\sum\limits_{n=0}^\infty \frac{\epsilon^{n}}{n+2} \\&=\frac{1}{\pi}(M \mathcal{k})^{\frac{2}{\alpha}}\epsilon+\Theta\left(\epsilon^2\right) \end{align}$$